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(D) . The equation is $y^2-2y+1 = -4x-3+1 \implies (y-1)^2 = -4(x+\frac{1}{2})$. Vertex is $\left(-\frac{1}{2}, 1\right)$,which is $(III)$.$(B)$. The

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$A-III, B-V, C-I, D-IV$

Vedclass · Answer

(D) . The equation is $y^2-2y+1 = -4x-3+1 \implies (y-1)^2 = -4(x+\frac{1}{2})$. Vertex is $\left(-\frac{1}{2}, 1\right)$,which is $(III)$.$(B)$. The equation is $x^2+8x+16 = -12y-4+16 \implies (x+4)^2 = -12(y-1)$. Vertex is $(-4, 1)$,which is $(V)$.$(C)$. The equation is $y^2-2y+1 = x-2+1 \implies (y-1)^2 = 1(x-1)$. Here $4a=1 \implies a=\frac{1}{4}$. Focus is $(h+a, k) = (1+\frac{1}{4}, 1) = \left(\frac{5}{4}, 1\right)$,which is $(I)$.$(D)$. The equation is $x^2-2x+1 = 8y+23+1 \implies (x-1)^2 = 8(y+3)$. Here $4a=8 \implies a=2$. Focus is $(h, k+a) = (1, -3+2) = (1, -1)$,which is $(IV)$.Thus,the correct match is $A-III, B-V, C-I, D-IV$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

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